Q. 80 - Precalculus Enhanced with Graphing Utilities

Question

Landscaping A landscape company is hired to plant trees in three new subdivisions. The company charges the developer for each tree planted, an hourly rate to plant the trees, and a fixed delivery charge. In one subdivision it took 166 labor hours to plant 250 trees for a cost of \(7520. In a second subdivision it took 124 labor hours to plant 200 trees for a cost of \)5945. In the final subdivision it took 200 labor hours to plant 300 trees for a cost of $8985. Determine the cost for each tree, the hourly labor charge, and the fixed delivery charge.

Step-by-Step Solution

Answer

The cost of each tree is $\$19.95,$ the hourly labor charge is $\$13.75,$ and the fixed delivery charge is $\$250.$

Step 1. Given Information

A company costs $7520  for one subdivision which took 166 labor hours to plant 250 trees, in a second subdivision it took 124 labor hours to plant 200 trees for a cost of $5945, and in the final subdivision it took 200 labor hours to plant 300 trees for a cost of $8985. We have to determine the cost for each tree, the hourly labor charge, and the fixed delivery charge. 

Step 2. Determining the cost for each tree, the hourly labor charge, and the fixed delivery charge

Let a be the cost for each tree, y be the hourly labor charge, and z be the fixed delivery charge. 

The system of equations we get are,

$\left\{\begin{array}{c}c+250a+166b=7520\\ c+200a+124b=5945\\ c+300a+200b=8985\end{array}\right.$

Step 3. Solve

Let's write the augmented matrix that represents the system

$$\begin{gathered} \left[\begin{array}{ccccc}1& 250& 166& |& 7520\\ 1& 200& 124& |& 5945\\ 1& 300& 200& |& 8985\end{array}\right] \\ {R}_2=r_1-r_2 \\ \left[\begin{array}{ccccc}1& 250& 166& |& 7520\\ 0& 50& 42& |& 1575\\ 1& 300& 200& |& 8985\end{array}\right] \\ {R}_2=\frac{1}{50}{r}_2 \\ \left[\begin{array}{ccccc}1& 250& 166& |& 7520\\ 0& 1& \raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$25$}\right.& |& \raisebox{1ex}{$63$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 1& 300& 200& |& 8985\end{array}\right] \end{gathered}$$

Step 4. Solve

By proceeding with the calculation further

$$\begin{gathered} R_3=r_1-r_3 \\ \left[\begin{array}{ccccc}1& 250& 166& |& 7520\\ 0& 1& \raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$25$}\right.& |& \raisebox{1ex}{$63$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& -50& -34& |& -1465\end{array}\right] \\ {R}_3=50r_2+r_3 \\ \left[\begin{array}{ccccc}1& 250& 166& |& 7520\\ 0& 1& \raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$25$}\right.& |& \raisebox{1ex}{$63$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 0& 8& |& 110\end{array}\right] \\ {R}_3=\frac{1}{8}{r}_3 \\ \left[\begin{array}{ccccc}1& 250& 166& |& 7520\\ 0& 1& \raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$25$}\right.& |& \raisebox{1ex}{$63$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 0& 1& |& \raisebox{1ex}{$55$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\end{array}\right] \end{gathered}$$

Step 5. Use row echelon form

The matrix is now in row echelon form. The final matrix represents the system

$\left\{\begin{array}{c}c+250a+166b=7520 \left(1\right)\\ a+\frac{21}{25}b=\frac{63}{2} \left(2\right)\\ b=\frac{55}{4}=13.75 \left(3\right)\end{array}\right.$

Step 6. Solve for a

To find a, we will substitute b in the equation in (2)

$$\begin{gathered} a+\frac{21}{25}\left(\frac{55}{4}\right)=\frac{63}{2} \\ a=\frac{63}{2}-\frac{231}{20} \\ a=\frac{399}{20} \\ a=19.95 \end{gathered}$$

Step 7. Solve for c

To find c, we will substitute a and b in the equation in (1)

$$\begin{gathered} c+250\left(19.95\right)+166\left(13.75\right)=7520 \\ c+4987.5+2282.5=7520 \\ c=250 \end{gathered}$$

Therefore, the cost of each tree is $\$19.95,$ the hourly labor charge is $\$13.75$, and the fixed delivery charge is $\$250.$